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THE MOLE!!!
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Ch 7New.notebook
October 31, 2012
23 representative particles of • a mole (mol) of a substance represents 6.02 X 10
that substance.
chemical • The mole is a unit of measurement for the amount of substance or amount. It is one of the base units in the International System of Units, and has the unit symbol mol
• What is means is that the mol is a word that stands for a number, like the word 23 dozen stands for the number 12. A mol of something has 6.02 X 10
representative particles of that substance.
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1) One billion Pb atoms is how many mols of Pb?
1,000,000,000 Pb atoms
1 X
1 mol Pb
= 1.66 X 10 ­15 mol Pb
23
6.02 X 1023 atoms Pb
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2) How many Hydrogen atoms are in 0.5 moles of sucrose (C12 H22 O11)
0.5 mol C12H22O11
1
x
6.02 X 1023 molecules C12H22O11
1 mol C12H22O11
22 atoms of H
x 1 molecule C
H O
12
22
11
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EX: What is the molar mass of the following:
1)Pb
2)NaCl
3)Water
4)Calcium chloride
5)C12 H22 O11
6)
Tin (IV) phosphate
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EX: What is the molar mass of the following:
1)Pb=207.2
2)NaCl23+35.45=58.45
3)WaterH2O 1+1+16=18
4)Calcium chloride CaCl2 40+2(35.45) =110.9
5)C12 H22 O11 12(12)+22(1)+11(16)=342
6)
Tin (IV) phosphate Sn3(PO4)4
3(118.7)+4(30.97)+16(16)=735.98
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The molar mass of something is the mass of 1.0 mol of that stuff in grams; just like adding up the atomic masses gives us the mass of just one of them in amu’s. That makes the unit of molar mass = g/mol
(molar mass)
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Name each of the following chemical compounds and list their molar masses to the nearest g/mol:
1) PbSO4
2) N2O3
3) CoCl2 . 4 H2O
4) B2F6
5) Sn(CO3)2
Write the formulas of each of the following chemical compounds and list their molar masses to the nearest g/mol:
6) copper (I) oxide
7) ammonium phosphate
8) vanadium (V) cyanide
9) nitrogen tribromide
10) iron (II) fluoride tetrahydrate
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Name each of the following chemical compounds and list their molar masses to the nearest g/mol:
1) PbSO4 lead (II) sulfate303 g/mol
2) N2O3 dinitrogen trioxide76 g/mol
3) CoCl2 . 4 H2O cobalt (II) chloride tetrahydrate 202 g/mol
4) B2F6 diboron hexafluoride136 g/mol
5) Sn(CO3)2 tin (IV) carbonate239 g/mol
Write the formulas of each of the following chemical compounds and list their molar masses to the nearest g/mol:
6) copper (I) oxide Cu2O143 g/mol
7) ammonium phosphate (NH4)3PO4 149 g/mol
8) vanadium (V) cyanide V(CN)5 181 g/mol
9) nitrogen tribromide NBr3 254 g/mol
10) iron (II) fluoride tetrahydrate FeF2 . 4 H2O166 g/mol
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Ch 7New.notebook
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Review of Day 1
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
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Ch 7New.notebook
October 31, 2012
Review of Day 1
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
So for water (for example):
6.02 X 1023 H2O molecules = 1 mol and 1 mol H2O = 18 g H2O
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Ch 7New.notebook
October 31, 2012
So... the molar mass of tin (IV) phosphate is 735.98 g/mol.
That means 1 mol of tin (IV) phosphate has a mass of 735.98 g
...and if you have 735.98 g of tin (IV) phosphate you have 1 mol
In dimensional analysis terms:
735.98 g Sn3(PO4)4
1 mol Sn3(PO4)4
both are legit conversion factors!
(PO4)4
735.98 g Sn3(PO4)4
1 mol Sn3
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Ch 7New.notebook
October 31, 2012
Review of Day 1
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
So for water (for example):
6.02 X 1023 H2O molecules = 1 mol and 1 mol H2O = 18 g H2O
To that lets add one more tidbit: 1 mol of any gas at STP has a volume of 22.4 L
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How many mols of water are in 47.9 g of water?
How many molecules of water is that?
How many molecules of water are in 47.9 g of water? (the above in one step)
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One mole of X is:
• 6.02 X 10 23 of X (in number)
• Has a mass in grams = the molar mass of X • If X is a gas at STP has a volume = 22.4 L
(grams)
October 31, 2012
(L)
• atoms
• molecules
• ions
• formula units
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Convert the following to mols:
1) 12. 8 g of sodium acetate
2) 8.0 L of dinitogen pentoxide gas at STP
3) 1.7 X1019 water molecules
How many molecules and what is the mass of 150.0 mL of carbon dioxide at STP?
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1) Convert the following to moles:
a) 1.6 L of dinitrogen pentoxide gas at STP
b) 43.8 g of copper metal
c) 2.87 X 10 23 atoms of lead
d) 19.4 g of calcium carbonate
e) 45,235 ml of neon gas at STP
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2) How many carbon dioxide molecules are in 6.70 g of carbon dioxide?
3) How many L of volume would 5.6 g of nitrogen gas (N2) at STP occupy?
4) If every cat in the world (11.2 billion of the beasts) was reduced to 1 carbon atom per cat (thus 11.2 billion C atoms) how many grams of carbon would they make?
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5) How many ammonium ions are in .0000536 g of ammonium phosphate?
6) What is the mass of air (roughly 79% N2 and 21% O2 ) in a 55 gallon drum at STP? (1 gal = 3.785 L)(even though there is no “air” molecule air can be given an average molecular mass based on it’s percent composition)
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Percent Composition
• Percent composition tells us the percent by mass of each element within an element
­We calculate this by: grams of the element X 100 % grams of the compound
if we’re not given grams we use 1 mol of the stuff so we can use the molar mass
EX: What is the % composition of nitric acid?
1) What is nitric acid? HNO3
2) What is its molar mass? 1 + 14 + 3(16) = 63 g/mol
3) Calculate: % H = 1/63 = .0159 = 1.59 % H
% N = 14/63 = .222 = 22.2 % N
% O = [3(16)] / 63 = .762 = 76.2 % O
we could check by adding up and getting close to 100%
What is the % composition of calcium acetate?
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Ch 7New.notebook
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Percent Composition
• Percent composition tells us the percent by mass of each element within an element
­We calculate this by: grams of the element X 100 % grams of the compound
if we’re not given grams we use 1 mol of the stuff so we can use the molar mass
EX: What is the % composition of nitric acid?
1) What is nitric acid? HNO3
2) What is its molar mass? 1 + 14 + 3(16) = 63 g/mol
3) Calculate: % H = 1/63 = .0159 = 1.59 % H
% N = 14/63 = .222 = 22.2 % N
% O = [3(16)] / 63 = .762 = 76.2 % O
we could check by adding up and getting close to 100%
What is the % composition of calcium acetate?
% Ca = 40/158 = .253 = 25.3%
% C = 48/158 = .304 = 30.4%
% H = 6/158 = .038 = 3.8%
% O = 64/158 = .405 = 40.5%
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• Another way we find % composition is by experiment
In an experiment 0.500 g of an acid was analyzed. It was found to be 0.010 g H, 0.163 g S, and 0.327 g O. What is the acid’s % composition?
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• Another way we find % composition is by experiment
In an experiment 0.500 g of an acid was analyzed. It was found to be 0.010 g H, 0.163 g S, and 0.327 g O. What is the acid’s % composition?
% H = .01/.5 = .02 = 2%
% S = .163/.5 = .326 = 32.6%
% O = .327/.5 = .654 = 65.4%
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Determining Empirical and Molecular formulas
• Before we go into the math of this we have to define these two things:
• Empirical formula: the lowest whole number ratio of the atoms in a compound
• Molecular formula: the actual number of atoms in the compound
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­ We can use % composition data to find empirical formulas
EX: What is the empirical formula of a compound that is 79.8 % C, and 20.2 % H 1) assume you have 100 g of the compound
that means you would have 79.8 g of C and 20.2 g of H
2) convert these to moles
79.8 g of C 1 mole of C = 6.65 mol
1 X 12 g of C of C
20.2 g of H 1 mole of H = 20.2 mol
1 X 1 g of H of H
3) divide by the smallest of the mole numbers
6.65 mol of C = 1 mol of C
6.65
20.2 mol of H = 3 mol of H
6.65
therefore the EF is: CH3
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Ch 7New.notebook
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what if we don’t get whole numbers when we divide?
• if it’s close to a whole number it probably is that number (4.06 = 4)
• all the numbers may need multiplied by a whole number to become whole.
EX : C = 2.3, H = 3, N = 1 X 3 to get C 7H9N3
Try finding the empirical formula for the acid in the previous part
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Ch 7New.notebook
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what if we don’t get whole numbers when we divide?
• if it’s close to a whole number it probably is that number (4.06 = 4)
• all the numbers may need multiplied by a whole number to become whole.
EX : C = 2.3, H = 3, N = 1 X 3 to get C 7H9N3
Try finding the empirical formula for the acid in the previous part
H = 2%, S = 32.6%, O = 65.4%
2 g H 1 mol H 1 1 g H 2 mol H / 1.017 = 1.96 2
32.6 g S 1 mol S 1.017 mol S / 1.017 = 1
1 32.06 g S
65.4 g O 1 mol O 4.088 mol O / 1.017 = 4.02
1 16 g O
4
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Finding molecular formulas:
• To find molecular formulas(MF) from empirical formulas(EF) we need another piece of information. The compounds molecular mass(MM)
• If we found a compounds EF to be CH 3 and we also know its MM to be 30
Add up the weight of the EF CH 3 = 15
Find how many times that goes into the compounds MM 30/15 = 2
Multiply the EF through by that number to get the MF CH3 X 2 = C2H6
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Summary Problem:
1,6 diaminohexane is used to make nylon. Analysis of this compound finds its percent composition to be 62.1% C, 13.8% H, and 24.1% N. Its molecular mass is also found to be 116 g/mol. What are the empirical and molecular formulas of this compound?
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Ch 7New.notebook
October 31, 2012
Summary Problem:
1,6 diaminohexane is used to make nylon. Analysis of this compound finds its percent composition to be 62.1% C, 13.8% H, and 24.1% N. Its molecular mass is also found to be 116 g/mol. What are the empirical and molecular formulas of this compound?
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Attachments
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